The Henderson–Hasselbalch Equation

A buffer's pH depends on the ratio of base to acid. Rearranging $K_a$ and taking logarithms gives a formula that shows this directly.

Key idea

$$\text{pH} = \text{p}K_a + \log\frac{[\mathrm{A^-}]}{[\mathrm{HA}]}$$ When the two are equal, $\text{pH} = \text{p}K_a$. Every factor of 10 in the ratio moves the pH by 1.

Why it works

Start from $K_a = \dfrac{[\mathrm{H_3O^+}][\mathrm{A^-}]}{[\mathrm{HA}]}$ and solve for $[\mathrm{H_3O^+}] = K_a\dfrac{[\mathrm{HA}]}{[\mathrm{A^-}]}$. Take $-\log$ of both sides: $-\log[\mathrm{H_3O^+}] = -\log K_a - \log\dfrac{[\mathrm{HA}]}{[\mathrm{A^-}]}$, that is, $\text{pH} = \text{p}K_a + \log\dfrac{[\mathrm{A^-}]}{[\mathrm{HA}]}$. In a buffer both are present in large amounts, so so little ionizes that you can use the amounts you mixed. Since both share the same volume, a ratio of moles works as well as a ratio of molarities.

The fractions of HA and A⁻ against pH. They cross at pH = pKa, and one unit away the ratio is 10 to 1.

Worked examples

Worked example: pH of an acetate buffer

A buffer has $[\mathrm{CH_3COO^-}] = 0.15\ \text{M}$ and $[\mathrm{CH_3COOH}] = 0.10\ \text{M}$ ($\text{p}K_a = 4.74$). Find its pH.

Ratio: $\dfrac{0.15}{0.10} = 1.5$; $\log 1.5 = 0.18$.
$\text{pH} = 4.74 + 0.18 = 4.92$. More base than acid, so the pH is above the $\text{p}K_a$, as it should be.

Answer: $\text{pH} = 4.92$.

Worked example: Designing a buffer

What ratio $[\mathrm{CH_3COO^-}]/[\mathrm{CH_3COOH}]$ gives pH 5.00?

$5.00 = 4.74 + \log(\text{ratio})$, so $\log(\text{ratio}) = 0.26$.
$\text{ratio} = 10^{0.26} = 1.82$.

Answer: $1.82$.

Worked example: A base buffer

An ammonia buffer has $[\mathrm{NH_3}] = 0.10\ \text{M}$ and $[\mathrm{NH_4^+}] = 0.20\ \text{M}$. $K_b(\mathrm{NH_3}) = 1.8 \times 10^{-5}$. Find the pH.

The acid is $\mathrm{NH_4^+}$: $\text{p}K_a = 14.00 - \text{p}K_b = 14.00 - 4.74 = 9.26$ (using $K_a = 5.6 \times 10^{-10}$ gives 9.25).
$\text{pH} = 9.25 + \log\dfrac{0.10}{0.20} = 9.25 - 0.30 = 8.95$.

Answer: $\text{pH} = 8.95$.

For a base buffer you can work with $K_b$ instead: $\text{pOH} = \text{p}K_b + \log\dfrac{[\mathrm{HB^+}]}{[\mathrm{B}]}$, then $\text{pH} = 14.00 - \text{pOH}$. It gives the same answer as using the $\text{p}K_a$ of the conjugate acid, which is usually quicker. Read the other way, the equation says the $\text{p}K_a$ of an acid is the pH at which it is exactly half ionized, which is how $\text{p}K_a$ values are measured.

Common mistake

Putting the acid over the base inside the log, which flips the sign of the correction. Check: more base than acid must give a pH above the $\text{p}K_a$; more acid, below. Remember "base over acid," as in the order the reaction makes them.

The equation works well when both concentrations are at least about 100 times $K_a$ and the ratio is between 0.1 and 10. Outside those limits, solve the equilibrium in full.

The Henderson–Hasselbalch Equation

Worked examples

Practice

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